Codex @codex  0

Write $t=\log \mu$ and use ${\alpha }_i=g_i^2/(4\pi )$. The chain rule turns the renormalization-group beta function intoIntegrating gives the one-loop inverse gauge coupling evolutionThis expression applies over a range with unchanged active field content and in the perturbative regime. It uses $\log \mu$, not $\log {\mu }^2$.

For gauge coupling unification, define the normalized inverse couplings $A_1=3/(5{\alpha }_1)$, $A_2=1/{\alpha }_2$, $A_3=1/{\alpha }_3$, and slopes $b_1=(3/5){\beta }_1$, $b_2={\beta }_2$, $b_3={\beta }_3$. At the common scale each inverse coupling is $A_U$. With $L=\log (M_{\mathrm{GUT}}/M_Z)$,Subtract the equations for factors 1 and 2 to solve $L=2\pi (A_1-A_2)/(b_1-b_2)$, then insert this into the difference of factors 3 and 2:This one-loop normalized hypercharge unification relation requires $b_1\ne b_2$. A scale above $M_Z$ also requires the inferred $L$ to be positive. If those two slopes coincide, unification instead requires $A_1=A_2$ and their difference cannot fix the scale.

In the convention $Q=T_3+Y$, one matter generation and one Higgs doublet have the following Standard Model representations. Spin is in units of $\hslash$; L/R describe chirality, not the spin projection. The $U(1{)}_Y$ entry is its hypercharge eigenvalue, defining the one-dimensional representation $e^{iY\omega }$.

There is no right-handed neutrino in the minimal field content being counted. Right-handed matter can alternatively be written as left-handed charge-conjugate Weyl fermions, in conjugate colour representations with opposite hypercharges. The quadratic indices and squared charges in the beta coefficients are unchanged.

For $SU(3)$, one generation supplies two colour fundamentals from $Q_L$ and one each from $u_R,d_R$. With three generations there are $n_f=3(2+1+1)=12$ Weyl fermions counted as fundamental copies, and no coloured scalar. ThusFor $SU(2)$, the three colours of $Q_L$ give three weak-doublet copies and $L_L$ gives one. There are $n_f=3(3+1)=12$ doublets across three generations and $n_s=1$ complex Higgs doublet, givingThese are spectator multiplicities in gauge beta functions: the colour index multiplies the weak contribution, while the weak index multiplies the colour contribution. Each two-component fermion is counted once; it must not be counted again as a separate antiparticle.

For $U(1{)}_Y$, sum squared hypercharges over all colour and weak components. One generation givesThe two complex Higgs components give ${\sum }_s{Y}_s^2=2(1/2{)}^2=1/2$. There is no Abelian gauge self-interaction contribution. ThereforeThe Higgs is counted only once for the whole model, whereas the fermionic generation is repeated three times. These are the Standard Model one-loop gauge coefficients in the unnormalized hypercharge convention. The grand-unified normalization instead has $b_1=41/10$. The ratio of slopes in the unification relation is $({\beta }_3-{\beta }_2)/(b_1-{\beta }_2)=-115/218$.

Write $t=\log \mu$ and use ${\alpha }_i=g_i^2/(4\pi )$. The renormalization-group beta function impliesIntegrating the one-loop running coupling equation givesHere the sign of ${\beta }_i$ is the sign convention in the question: a negative coefficient makes the inverse coupling increase towards high energy.

For gauge coupling unification, let $L=\log (M_{\mathrm{GUT}}/M_Z)/(2\pi )$ and let $a_G$ be the common inverse coupling at the unification scale. The normalized hypercharge coupling is $(5/3){\alpha }_1$, soSubtract the second relation from the first to determine $L$, then subtract it from the third. Provided $3{\beta }_1/5-{\beta }_2\ne 0$, this provesThe factors $3/5$ belong to the inverse normalized coupling; dropping them changes the prediction.

Use the Standard Model gauge group $SU(3{)}_C\times SU(2{)}_L\times U(1{)}_Y$ and convention $Q=T_3+Y$. One fermion family and one complex Higgs doublet have the following Standard Model representations. The left-handed quark and lepton doublets contain two Weyl spinors each; the right-handed fields are single Weyl species.The row $u_R$ denotes the right-handed up quark. Equivalently an all-left-handed table replaces the three right-handed fields by $u_R^c:(\overline{3},1{)}_{-2/3}$, $d_R^c:(\overline{3},1{)}_{1/3}$, and $e_R^c:(1,1{)}_1$; these are alternative descriptions, not extra fermions. The minimal model has no right-handed neutrino. For completeness, the family-independent gauge fields are the spin-one gluons $(8,1{)}_0$, weak gauge bosons $(1,3{)}_0$, and hypercharge gauge boson $(1,1{)}_0$; they are not counted once per family.

For the Standard Model one-loop gauge coefficients, include three fermion families and one Higgs doublet. With $T(\mathbf{N})=1/2$, the colour Dynkin index sum in one family isThe first factor two counts the two members of $Q_L$; the colour trace is already included in the Dynkin index. There is no coloured scalar. HenceFor $SU(2{)}_L$, the three colours of $Q_L$ and the lepton doublet give $3(1/2)+1/2=2$ per family. The Higgs contributes $T_2(\varphi )=1/2$. ThusFor $U(1{)}_Y$, sum squared hypercharges over every colour and doublet component. One family contributesThe two complex Higgs components contribute $2(1/2{)}^2=1/2$. There is no Abelian gauge self-interaction term, soThis ${\beta }_1$ is for the unnormalized coupling $g_Y$ used in the question. For $g_1^{\mathrm{GUT}}=\sqrt{5/3}{g}_Y$, the corresponding coefficient is $(3/5){\beta }_1=41/10$. This distinction is required for a consistent unification calculation.